Answer:
d. 6.0 m
Explanation:
Given;
initial velocity of the car, u = 7.0 m/s
distance traveled by the car, d = 1.5 m
Assuming the car to be decelerating at a constant rate when the brakes were applied;
v² = u² + 2(-a)s
v² = u² - 2as
where;
v is the final velocity of the car when it stops
0 = u² - 2as
2as = u²
a = u² / 2s
a = (7)² / (2 x 1.5)
a = 16.333 m/s
When the velocity is 14 m/s
v² = u² - 2as
0 = u² - 2as
2as = u²
s = u² / 2a
s = (14)² / (2 x 16.333)
s = 6.0 m
Therefore, If the car had been moving at 14 m/s, it would have traveled 6.0 m before stopping.
The correct option is d
Answer:
F = - K x
a) K = 1.3 kg * 9.8 m/s^2 / .096 m = 133 kg/sec^2
b) ω = (K/m)^1/2 angular frequency of SHM
ω = (133 / 1.3)^1/2 = 10.1 / sec
f = 2 π ω = 6.28 * 10.1 / sec = 63.5 / sec
P = 1/f = .0157 sec
The distance quantity/ measurement must be squared.
Answer:
Stretch can be obtained using the Elastic potential energy formula.
The expression to find the stretch (x) is 
Explanation:
Given:
Elastic potential energy (EPE) of the spring mass system and the spring constant (k) are given.
To find: Elongation in the spring (x).
We can find the elongation or stretch of the spring using the formula for Elastic Potential Energy (EPE).
The formula to find EPE is given as:

Rewriting the above expression in terms of 'x', we get:

Example:
If EPE = 100 J and spring constant, k = 2 N/m.
Elongation or stretch is given as:

Therefore, the stretch in the spring is 10 m.
So, stretch in the spring can be calculated using the formula for Elastic Potential Energy.